幾道積分練習題 | 微積分福音

\begin{align*} \int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\,\mathrm{d}x \end{align*}

\begin{align} &\,\int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\,\mathrm{d}x\\[4mm] =&\,\frac{1}{\sqrt{2}}\int\frac{\sin(x)\cos(x)}{\sin(x+\frac{\pi}{4})}\,\mathrm{d}x\\[4mm] =&\,\frac{1}{2\sqrt{2}}\frac{\Big[\sin(u)-\cos(u)\Big]\Big[\cos(u)+\sin(u)\Big]}{\sin(u)}\mathrm{d}u\\[4mm] =&\,\frac{1}{2\sqrt{2}}\int\frac{2\sin^2(u)-1}{\sin(u)}\mathrm{d}u\\[4mm] =&\,\frac{1}{\sqrt{2}}\int\sin(u)\mathrm{d}u -\frac{1}{2\sqrt{2}}\int\csc(u)\mathrm{d}u\\[4mm] =&\,-\frac{\cos(x+\frac{\pi}{4})}{\sqrt{2}}\\[4mm] &\,\;-\frac{1}{2\sqrt{2}}\ln\big\vert\csc(x+\frac{\pi}{4}) -\cot(x+\frac{\pi}{4})}\big\vert+C \end{align}

\begin{align} &\,\int\frac{\sin(x)\cos(x)}{\sin(x)+\cos(x)}\,\mathrm{d}x\\[4mm] =&\,\frac{1}{2}\int\frac{\big(\sin(x)+\cos(x)\big)^2-1}{\sin(x)+\cos(x)}\,\mathrm{d}x\\[4mm] =&\,\frac{1}{2}\int\big(\sin(x)+\cos(x)\big)\,\mathrm{d}x\\[4mm] &\,\quad\frac{1}{2\sqrt{2}}\int\frac{1}{\sin(x+\frac{\pi}{4})}\,\mathrm{d}x\\[4mm] =&\,\frac{1}{2}\big(-\cos(x)+\sin(x)\big)\\[4mm] &\,\quad-\frac{1}{2\sqrt{2}}\ln\abs{\csc(x+\frac{\pi}{4}) -\cot(x+\frac{\pi}{4})}+C \end{align}

\begin{align*} \int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}\,\mathrm{d}x \end{align*}

由於被積分函數是指數函數乘上一坨東西，所以猜測答案的形式亦是

\begin{align} e^{\sin(x)}g(x) \end{align}

於是

\begin{align*} &\,\Big(e^{\sin(x)}g(x)\Big)'\\[4mm] =&\,e^{\sin(x)}\cos(x)g(x)+e^{\sin(x)}g'(x)\\[4mm] =&\,e^{\sin(x)}\Big[\cos(x)g(x)+g'(x)\Big] \end{align*}

接下來解

\begin{align*} &\,\cos(x)g(x)+g'(x)\\[4mm] =&\,\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}\\[4mm] =&\,x\cos(x)-\tan(x)\sec(x) \end{align*}

$g(x)=x$ 是明顯說不通的，所以無中生有一下

\begin{align} &\,\cos(x)g(x)+g'(x)\\[4mm] =&\,x\cos(x)-1+1-\tan(x)\sec(x)\\[4mm] =&\,\cos(x)\Big[x-\sec(x)\Big]+\Big[1-\tan(x)\sec(x)\Big] \end{align}

成功地求出 $g(x)=x-\sec(x)$ ，故答案為 $e^{\sin(x)}\Big(x-\sec(x)\Big)+C$ 。

\begin{align} &\,\int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}\,\mathrm{d}x\\[4mm] =&\,\int xe^{\sin(x)}\cos(x)\,\mathrm{d}x\\[4mm] &\,\quad-\int e^{\sin(x)} \tan(x)\sec(x)\,\mathrm{d}x\\[4mm] =&\,xe^{\sin(x)}-\int e^{\sin(x)}\,\mathrm{d}x\\[4mm] &\,\quad-e^{\sin(x)}\sec(x) +\int e^{\sin(x)}\,\mathrm{d}x\\[4mm] =&\,e^{\sin(x)}\big(x-\sec(x)\big)+C \end{align}

\int\frac{\ln(x)}{(1+x^2)^{\frac{3}{2}}}\,\mathrm{d}x

設 $x=\sinh(t) , \,\mathrm{d}x=\cosh(t)\mathrm{d}t$ ，則

\begin{align} &\,\int \frac{\ln\big(\sinh(t)\big)}{\cosh^3(t)}\cosh(t)\mathrm{d}t\\[4mm] =&\,\int\mathrm{sech}^2(t)\ln\big(\sinh(t)\big)\mathrm{d}t\\[4mm] =&\,\tanh(t)\ln\big(\sinh(t)\big)\\[4mm] &\quad-\int\underbrace{\tanh(t)\cdot\frac{\cosh(t)}{\sinh(t)}}_{=1}\mathrm{d}t\\[4mm] =&\,\frac{\sinh(t)\ln\big(\sinh(t)\big)}{\sqrt{1+\sinh^2(t)}}+t+C\\[4mm] =&\,\frac{x\ln(x)}{\sqrt{1+x^2}}+\sinh^{-1}(x)+C \end{align}

設 $x=\frac{1}{u} , \,\mathrm{d}x=-\frac{\mathrm{d}u}{u^2}$ ，則

\begin{align} &\,\int\frac{u\ln(u)}{(1+u^2)^{\frac{3}{2}}}\mathrm{d}u\\[4mm] =&\,-\frac{\ln(u)}{\sqrt{1+u^2}} +\int\frac{\mathrm{d}u}{u\sqrt{1+u^2}}\\[4mm] =&\,\frac{x\ln(x)}{\sqrt{1+x^2}} -\int\frac{\,\mathrm{d}x}{\sqrt{1+x^2}}\\[4mm] =&\,\frac{x\ln(x)}{\sqrt{1+x^2}}+\sinh^{-1}(x)+C \end{align}